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3 years ago

10

9) One line of text on a page uses about 3/16 of an inch. There are 1-inch 10 margins at the top and bottom of a page. Write and

solve an inequality to find the number of lines that can be typed on a page that is 11 inches long.

1 answer:

arsen [322]3 years ago

8 0

11-2(1)
This is for the top and bottom margins. You are left with 9. Divide that by 3/16 or 0.1875 to get 48.
(11-2)/0.1875

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3x to the power of 2 - 363 = 0

Vladimir79 [104]

Answer:

yes

Step-by-step explanation:

because it would be 300 so u would be in negative so ya

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3 years ago

What is the answer of 7 copies of the sum of 8/5 + 4

kotykmax [81]

To solve (8/5 + 4)7, you first have to add 8/5+4. But you cannot simply add the too because they have to be in a common denominator 'form'. To do that, we find a common denominator which is 5. We then multiply 4/1 by 5.
8/5 + 4/1= 8/5 +20/5.
Now you can solve; 8+20= 28. You leave the denominator as 5.
You get 28/5 and now you have to multiply it by 7/1. Another of saying 28/5 is 5 3/5. So 5 3/5 x 7/1= 39 2/10 (final answer)

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4 years ago

How many permutations of the 26 letters of the English alphabet do not contain any of the strings fish, rat, or bird

NARA [144]

The number of permutations of the 26 letters of the English alphabet that do not contain any of the strings fish, rat, or bird is 402619359782336797900800000

Let

$\mathcal{E}=\{\text{All lowercase letters of the English Alphabet}\}\\\\B=\overline{\{b,i,r,d\}} \cup \{bird\}\\\\F=\overline{\{f,i,s,h\}} \cup \{fish\}\\\\R=\overline{\{r,a,t\}} \cup \{rat\}\\\\FR=\overline{\{f,i,s,h,r,a,t\}} \cup \{fish,rat\}$

Then

$Perm(\mathcal{E})=\{\text{All orderings of all the elements of } \mathcal{E}\}\\\\Perm(B)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing bird}\}\\\\Perm(F)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing fish}\}\\\\Perm(R)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing rat}\}\\\\Perm(FR)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing both fish and rat}\}\\$

Note that since

$F \cap R=\varnothing$ , $Perm(F)\cap Perm(R)\ne \varnothing$

But since

$B \cap R \ne \varnothing$ , $Perm(B)\cap Perm(R)= \varnothing$

and

$B \cap F \ne \varnothing$ , $Perm(B)\cap Perm(F)= \varnothing$

Since

$|\mathcal{E} |=26 \text{, then, } |Perm(\mathcal{E})|=26! \\\\|B|=26-4+1=23 \text{, then, } |Perm(B)|=23!\\\\|F|=26-4+1=23 \text{, then, } |Perm(F)|=23!\\\\|R|=26-3+1=24 \text{, then, } |Perm(R)|=24!\\\\|FR|=26-7+2=21 \text{, then, } |Perm(FR)|=21!\\$

where $|Perm(X)|=\text{number of possible permutations of the elements of X taking all at once}$

and

$|Perm(F) \cup Perm(R)| = |Perm(F)| + |Perm(R)| - |Perm(FR)|\\= 23!+24!- 21! \text{ possibilities}$

What we are looking for is the number of permutations of the 26 letters of the alphabet that do not contain the strings fish, rat or bird, or

$|Perm(\mathcal{E})|-|Perm(B)|-|Perm(F)\cup Perm(R)|\\= 26!-23!-(23!+24!- 21!)\\= 402619359782336797900800000 \text{ possibilities}$

This link contains another solved problem on permutations:

brainly.com/question/7951365

6 0

3 years ago

What is 7.03 as a mixed number or fraction in simplest form

Alona [7]

7.03 can be converted to 7 3/100.

.03 is the equivalent to 3 hundredths, we know this because it in the hundredths place.

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3 years ago

Please help!<br> This is mathematics!

Elanso [62]

Answer:

30, 30, 32, 32, 34, 38, 38, 38. Thats how you first need to set it up then what you do is add them all up which would equal to 272. Then you take 272 dived by how many numbers there are (theres 8 numbers) so you do 272/8 which would be 34! So 34 would be your answer.

Step-by-step explanation:

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4 years ago

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